[tex]\bf 19sin(x)+9=sin(x)\implies 19sin(x)-sin(x)=-9
\\\\\\
18sin(x)=-9\implies sin(x)=\cfrac{-9}{18}\implies sin(x)=-\cfrac{1}{2}
\\\\\\
sin^{-1}[sin(x)]=sin^{-1}\left( -\cfrac{1}{2} \right)\implies \measuredangle x=sin^{-1}\left( -\cfrac{1}{2} \right)
\\\\\\
\measuredangle x =
\begin{cases}
\frac{7\pi }{6}\\\\
\frac{11\pi }{6}
\end{cases}[/tex]